What is the maximum volume in cubic inches of an open box to be made from a 10-inch by 20-inch piece of cardboard by cutting out squares of equal sides from the four corners and bending up the sides? Your work must include a statement of the function and its derivative.

Question

anonymous · Answer

@Lyrae @AlexandervonHumboldt2 @CausticSyndicalist @mathmate @ShadowLegendX @SolomonZelman @TheSmartOne @sleepyjess

mathmate · Answer

Have you started calculus yet?

anonymous · Answer

yes

anonymous · Answer

i think the dimensions are V=(20-2x)(10-2x)x

mathmate · Answer

|dw:1427243900775:dw|
Yes your expression is correct!
V(x)=x(20-2x)(10-2x)

mathmate · Answer

You will need to find V'(x), equate to zero and solve for x.
One of the solutions, say x1 will be minimum,and the other will (x2) be a  maximum.
To distinguish between the two, you calculate
V"(x).  V(x1)>0 => minimum, and V(x2)<0 => maximum.

mathmate · Answer

* V"(x1)>0 => minimum, V"(x2)<0 => maximum.

anonymous · Answer

$$f'(x)=4x(x^2-15x+50)$$

anonymous · Answer

@mathmate

mathmate · Answer

V(x)=x(20-2x)(10-2x)
=$4x^3-60x^2+200x$
Then you can take derivative

anonymous · Answer

$$12x^2-120x+200$$

anonymous · Answer

its the derivative so then? @mathmate

mathmate · Answer

Your V'(x)=12x2−120x+200 is correct.
Now, you need to find x by equating V'(x)=0, i.e. solve the quadratic to find the maximum/minimum.

anonymous · Answer

|dw:1427245137219:dw|